Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

A warehouse is being built that will have neither heating nor cooling. Depending on the amount of insulation, the time constant for the building may range from 1 to 5 hr. To illustrate the effect insulation will have on the temperature inside the warehouse, assume the outside temperature varies as a sine wave, with a minimum of $16 ° C$ at $2 : 00 a . m .$ and a maximum of $32 ° C$ at $2 : 00 p . m .$ Assuming the exponential term (which involves the initial temperature T0) has died off, what is the lowest temperature inside the building if the time constant is 1 hr? If it is 5 hr? What is the highest temperature inside the building if the time constant is 1 hr? If it is 5 hr?

If the time constant is 1 hour, the lowest temperature inside the building will reach $16.3 ° C$ and the highest temperature will reach $31.7 ° C$ .

If the time constant is 5 hours, the lowest temperature inside the building will reach $19.1 ° C$ and the highest temperature will reach $28 . 9^{o} C$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: Given data.

The temperature outside the building varies as a sine wave, with a minimum of $16 ° C$ at $2 : 00 a . m .$ and a maximum of $32 {}^{\circ}C$ at $2 : 00 p . m .$ If the time constants for the building are 1 hours and 5 hours, it has to find the lowest and highest temperatures inside the building.

Step 2: Analyzing the given statement

$M = M_{0} - B c o s (ω t)$ …… (1)

$M_{0} - B = 16 . . . . . . (a) M_{0} + B = 32 . . . . . . (b)$

At $2 : 00 a . m .$ t=0,

And at 2:00 p.m, t=12 hours

Adding the equations (a) and (b),

$2 M_{0} = 48 M_{0} = 24$

Therefore, B=8.

Step3: To determine the temperature at time t.

The forcing function $Q (t)$ is given by,

$Q (t) = K (M_{0} - B c o s (ω t)) Q (t) = K (24 - 8 c o s (ω t))$

Temperature T(t) is given by

$T (t) = B_{0} - B F (t) + C e^{- k t}$ …… (2)

Where, $F (t) = \frac{c o s (ω t) + (\frac{ω}{k}) s i n (ω t)}{1 + \frac{ω^{2}}{k^{2}}}$ .

Substituting K=1,B=8 and B₀ =M₀ =24 in equation (2),

$T (t) = 24 - 8 F (t) + C e^{- t}$

Now as the exponential term died off, therefore,

$T (t) = 24 - 8 F (t)$ …… (3)

Where, the value of F(t) is,

$F (t) = \frac{1}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}} (\frac{c o s (ω t)}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}} + \frac{(\frac{ω}{k}) s i n (ω t)}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}})$

$F (t) = \frac{s i n (ω t + t a n^{- 1} (\frac{k}{ω}))}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}}$ …… (4)

Step 4: To determine lowest and highest temperatures inside the building if the time constant is 1 hr

Now as the maximum value of sin x is 1.

Therefore, from equation (4),

$F (t) = \frac{1}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}}$

So, by substituting the value of F(t) in equation (3),

$T (t) = 24 - \frac{8}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}}$

…… (5)

When the time constant is 1 hour i.e., when $\frac{1}{k} = 1$

And $ω = \frac{π}{12}$

Therefore, from equation (5),

Let $T_{L}$ be the lowest temperature,

$T_{L} = 24 - \frac{8}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}} T_{L} = 24 - \frac{8}{\sqrt{1 + \frac{π^{2}}{144}}} T_{L} = 16 . 3^{0} C$

Now as the minimum value of sin x is 1.

Therefore, from equation (4),

$F (t) = - \frac{1}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}}$

Thus, from equation (5),

Let $T_{H}$ be the highest temperature,

$T_{H} = 24 + \frac{8}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}} T_{H} = 24 + \frac{8}{\sqrt{1 + \frac{π^{2}}{144}}} T_{H} = 31 . 7^{0} C$

Hence, if the time constant is 1 hour, the lowest temperature inside the building will reach role="math" localid="1664185402769" $16.3 ° C$ and the highest temperature will reach role="math" localid="1664185416044" $31.7 ° C$ .

Step 5: To determine lowest and highest temperatures inside the building if the time constant is 5 hr

When the time constant is 5 hours i.e., when $\frac{1}{K} = \frac{1}{5}$

Hence, from equation (5),

Let $T_{L}$ be the lowest temperature,

$T_{L} = 24 - \frac{8}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}} T_{L} = 24 - \frac{8}{\sqrt{1 + \frac{25 π^{2}}{144}}} T_{L} = 19 . 1^{0} C$

Let $T_{H}$ be the highest temperature,

So, from equation (5),

$T_{H} = 24 + \frac{8}{\sqrt{1 + \frac{ω^{2}}{k^{2}}}} T_{H} = 24 + \frac{8}{\sqrt{1 + \frac{25 π^{2}}{144}}} T_{H} = 28 . 9^{0} C$

Thereafter, if the time constant is 5 hours, the lowest temperature inside the building will reach $19.1 ° C$ and the highest temperature will reach $28.9 ° C$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step1: Given data.

Step 2: Analyzing the given statement

Step3: To determine the temperature at time t.

Step 4: To determine lowest and highest temperatures inside the building if the time constant is 1 hr

Step 5: To determine lowest and highest temperatures inside the building if the time constant is 5 hr

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Fundamentals Of Differential Equations And Boundary Value Problems

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Step1: Given data.

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Step3: To determine the temperature at time t.

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Step 5: To determine lowest and highest temperatures inside the building if the time constant is 5 hr

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